Cambridge A Level Mathematics 9709 — 2017 May/June Paper 3 · Variant 2

2 Solve the inequality x 3x [4] −3 < −4. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 EITHER: (B1 2 2 State or imply non-modular inequality ( x − 3) < (3 x − 4) , or corresponding equation Make reasonable attempt at solving a three term quadratic M1 7 A1 Obtain critical value x = 4 7 A1) State final answer x > only 4 OR1: (B1 State the relevant critical inequality 3 − x < 3 x − 4 , or corresponding equation Solve for x M1 7 A1 Obtain critical value x = 4 7 A1) State final answer x > only 4 OR2: (B1 Make recognizable sketches of y = x − 3 and y = 3 x − 4 on a single diagram Find x-coordinate of the intersection M1 7 A1 Obtain x = 4 7 A1) State final answer x > only 4 Total: 4

More questions on Quadratics

3 (i) Express the equation cot tan sin in the form a b c 0, where a, b and c are constants to be determined.1 −2 1 = 21 cos41 + cos21 + = [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Hence solve the equation cot tan sin for [2] 1 −2 1 = 21 90Å < 1 < 180Å. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) Use correct formulae to express the equation in terms of cos θand sin θ M1 Use Pythagoras and express the equation in terms of cos θonly M1 4 2 A1 Obtain correct 3-term equation, e.g. 2cos θ+ cos θ− 2 = 0 Total: 3 3(ii) 2 M1 Solve a 3-term quadratic in cos θfor cos θ Obtain answer θ= 152.1°only A1 Total: 2

More questions on Trigonometry

4 The parametric equations of a curve are x t2 1, y 4t ln 2t . = + = + −1 dy (i) Express in terms of t. [3] dx ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the equation of the normal to the curve at the point where t 1. Give your answer in the form ax by c 0. = [3] + + = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) dy 2 B1 State = 4 + dt 2t − 1 dy dy dx M1 Use = ÷ dx dt dt dy 8t − 2 2 2 A1 Obtain answer = , or equivalent e.g. + dx 2t (2t − 1) t 2 4t − 2t Total: 3 4(ii) Use correct method to find the gradient of the normal at t = 1 M1 Use a correct method to form an equation for the normal at t = 1 M1 Obtain final answer x + 3 y − 14 = 0 , or horizontal equivalent A1 Total: 3

More questions on Differentiation

5 In a certain chemical process a substance A reacts with and reduces a substance B. The masses of A dy and B at time t after the start of the process are x and y respectively. It is given that and dt = −0.2xy 10 x . At the beginning of the process y 100. 2 = 1 t = + (i) Form a differential equation in y and t, and solve this differential equation. 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(ii) Find the exact value approached by the mass of B as t becomes large. State what happens to the mass of A as t becomes large. 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Mark scheme: 5(i) d y 2 y B1 State = − , or equivalent d t (1 + t ) 2 Separate variables correctly and attempt integration of one side M1 Obtain term ln y , or equivalent A1 2 A1 Obtain term , or equivalent (1 + )t Use y = 100 and t = 0 to evaluate a constant, or as limits in an expression containing terms of M1 b the form a ln y and 1 + t 2 A1 Obtain correct solution in any form, e.g. ln y = − 2 + ln100 1 + t Total: 6 5(ii) 100 B1 State that the mass of B approaches , or exact equivalent e 2 State or imply that the mass of A tends to zero B1 Total: 2

More questions on Differential equations

6 Throughout this question the use of a calculator is not permitted. The complex number 2 is denoted by u. −i (i) It is given that u is a root of the equation x3 ax2 b 0, where the constants a and b are real. Find the values of a and b. + −3x + = [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying both the inequalities z 1 and z z i . [4] −u < < +

Mark scheme: 6(i) EITHER: (M1 2x 3x Substitute x = 2 – i (or x = 2 + i ) in the equation and attempt expansions of and Equate real and/or imaginary parts to zero M1 Obtain a = – 2 A1 Obtain b = 10 A1) OR1: (M1 2x 3x Substitute x = 2 – i in the equation and attempt expansions of and Substitute x = 2 + i in the equation and add/subtract the two equations M1 Obtain a = – 2 A1 Obtain b = 10 A1) OR2: (M1  2  Factorise to obtain ( x − 2 + i )( x −−2 i )( x − p )  = x − 4 x + 5 x − p )   ( )(  Compare coefficients M1 Obtain a = – 2 A1 Obtain b = 10 A1) OR3: (M1 2 Obtain the quadratic factor ( x − 4 x + 5 ) Use algebraic division to obtain a real linear factor of the form x − p and set the remainder M1 equal to zero Obtain a = – 2 A1 Obtain b = 10 A1) OR4: (M1 Use αβ = 5 and α+ β = 4 in αβ+ βγ + γα= − 3 Solve for γ and use in αβγ = −b and/or α+ β+ γ = −a M1 Obtain a = – 2 A1 Obtain b = 10 A1) OR5: (M1 Factorise as (x–- (2-i))(x2 + ex + g) and compare coefficients to form an equation in a and b Equate real and/or imaginary parts to zero M1 Obtain a = – 2 A1 Obtain b = 10 A1) Total: 4 6(ii) Show a circle with centre 2 − i in a relatively correct position B1 Show a circle with radius 1 and centre not at the origin B1 Show the perpendicular bisector of the line segment joining 0 to – i B1 Shade the correct region B1 Total: 4

More questions on Complex numbers

1 dy 7 (i) Prove that if y then sec tan [2] cos = = 1 1. 1 d1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 1 sin (ii) Prove the identity 2 2 sec tan [3] + 1 1  sec21 + 1 1 −1. −sin 1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 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........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 1 40 1 sin (iii) Hence find the exact value of [4] + 1 1 d1. 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Mark scheme: 7(i) Use quotient or chain rule M1 Obtain given answer correctly A1 Total: 2 7(ii) EITHER: (M1 Multiply numerator and denominator of LHS by 1 + sinθ Use Pythagoras and express LHS in terms of sec θand tanθ M1 Complete the proof A1) OR1: (M1 Express RHS in terms of cos θand sin θ Use Pythagoras and express RHS in terms of sin θ M1 Complete the proof A1) OR2: (M1 Express LHS in terms of secθ and tanθ Multiply numerator and denominator by secθ + tanθ and use Pythagoras M1 Complete the proof A1) Total: 3 7(iii) Use the identity and obtain integral 2tanθ+ 2secθ− θ B2 Use correct limits correctly in an integral containing terms a tanθand b secθ M1 1 A1 Obtain answer 2 2 − 4π Total: 4

More questions on Trigonometry

9 Relative to the origin O, the point A has position vector given by OA i 2j 4k. The line l has −−→ equation r 9i 8k 3i 2k . = + + = −j + + - −j + (i) Find the position vector of the foot of the perpendicular from A to l. Hence find the position vector of the reflection of A in l. 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(ii) Find the equation of the plane through the origin which contains l. Give your answer in the form ax by cz d. 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(iii) Find the exact value of the perpendicular distance of A from this plane. 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Mark scheme: 9(ii) EITHER: (B1 Use scalar product to obtain an equation in a, b and c, e.g. 3a − b + 2 c = 0 Form a second relevant equation, e.g. 9a – b + 8c = 0 and solve for one ratio, e.g. a : b M1 Obtain final answer a : b : c = 1 : 1 : – 1 and state plane equation x + y – z = 0 A1) OR1: (M1 Attempt to calculate vector product of two relevant vectors, e.g. (3i −+j 2k ) × (9 i −+j 8k ) Obtain two correct components A1 Obtain correct answer, e.g. −6 i − 6 j + 6k , and state plane equation −−x y + z = 0 A1) OR2: (M1 Using a relevant point and relevant vectors, attempt to form a 2-parameter equation for the r = 6i + 6k + s (3i −+j 2k ) + t (9i −+j 8k ) plane, e.g. State 3 correct equations in x, y, z, s and t A1 x + y − z = 0 A1) Eliminate s and t and state plane equation , or equivalent OR3: (M1 Using a relevant point and relevant vectors, attempt to form a determinant equation for the x − 3 y − 1 z − 4 plane, e.g. 3 −1 2 = 0 9 −1 8 Expand a correct determinant and obtain two correct cofactors A1 Obtain answer − 6 x − 6 y + 6 z = 0 , or equivalent A1) Total: 3 9(iii) EITHER: (M1 uuur Using the correct processes, divide the scalar product of OA and a normal to the plane by the modulus of the normal or make a recognisable attempt to apply the perpendicular formula 1 + 2 − 4 A1 FT Obtain a correct expression in any form, e.g. , or equivalent 2 2 2 (1 + 1 + ( −1) ) Obtain answer 1 3 , or exact equivalent A1) OR1: (B1 FT Obtain equation of the parallel plane through A, e.g. x + y – z = – 1 [The f.t. is on the plane found in part (ii).] Use correct method to find its distance from the origin M1 Obtain answer 1 3 , or exact equivalent A1) OR2: (B1 FT Form equation for the intersection of the perpendicular through A and the plane [FT on their n] Solve for λ M1 1 A1) λn = 3 Total: 3

More questions on Vectors

10 y M x O p 140 The diagram shows the curve y x2 cos 2x for 0 The curve has a maximum point at M where x p. = ≤x ≤140. = 1 @1 A (i) Show that p satisfies the equation p . [3] 2 tan−1 p = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ @ A 1 1 (ii) Use the iterative formula to determine the value of p correct to 2 decimal 2 tan−1 pn+1 = pn places. Give the result of each iteration to 4 decimal places. 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(iii) Find, showing all necessary working, the exact area of the region bounded by the curve and the x-axis. 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Mark scheme: 10(i) Use correct product rule M1 2 A1 Obtain correct derivative in any form y ′ = 2 x cos2 x − 2 x sin 2 x ( ) Equate to zero and derive the given equation A1 Total: 3 10(ii) Use the iterative formula correctly at least once e.g. M1 0.5 → 0.55357 → 0.53261 → 0.54070 → 0.53755 Obtain final answer 0.54 A1 Show sufficient iterations to 4 d.p. to justify 0.54 to 2 d.p., or show there is a sign change in A1 the interval (0.535, 0.545) Total: 3 10(iii) 2 *M1 Integrate by parts and reach ax sin 2 x + b ∫ x sin 2 x dx 1 2 1 A1 Obtain x sin 2 x −∫ 2 x. sin 2 x dx 2 2 1 2 1 1 A1 Complete integration and obtain x sin 2 x + x cos2 x − sin 2 x , or equivalent 2 2 4 1 DM1 Substitute limits x = 0, x = 4π , having integrated twice 1 2 A1 Obtain answer (π − 8) , or exact equivalent 32 Total: 5

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